Question

Suppose, $z_1,z_2,z_3$ are the vertices of an equilateral triangle inscribed in the circle $\left|z\right|=2.$ If $z_1=1+i\sqrt{3}$ then $z_2$ and $z_3$ are equal to

• A. $-2,1-i\sqrt{3}$
• B. $2,1-i\sqrt{3}$
• C. $-2,1+i\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $-2,1-i\sqrt{3}$

$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$ lie on $\left|z\right|=2$ whose centre is at $O(0,0)$ and radius at $2$.

$z_1=1+i\sqrt{3}$ hence $\left|z\right|=2$ and

$arg\left(z_1\right)=\frac{\pi }{3}$

In turn $\left|z_2\right|=\left|z_3\right|=2$ and $arg\left(z_2\right)=arg\left(z_1\right)+{120}^0={180}^0$

$\therefore z_2=-2$

Further, $arg\left(z_3\right)=arg\left(z_2\right)+{120}^0={300}^0$

Hence, $z_3=2[\cos (2\pi -\frac{\pi }{3})+i\sin (2\pi -\frac{\pi }{3})]$

$=2[\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}]=2(\frac{1}{2}-\frac{\sqrt{3}i}{2})=1-\sqrt{3}i$

Thus, $z_2=-2$ and $z_3=1-i\sqrt{3}$